Hello!
Unfortunately you have no drawings attached, but I can envision.
I suppose all three corrals has one common dimension, denote its length as `x.` Then there are 4 walls with this length and 2 walls with another length, denote it as `y.`
So we have to use `4x+2y=2000` (ft) of fence while the enclosed area to maximize is `x*y.`
Express `y` from the equation, `y=1000-2x,` and substitute it into the formula for the area:
`A=A(x)=x*(1000-2x)=-2x^2+1000x.`
Of course `xgt=0` and `ygt=0,` so `xlt=500.`
The graph of this function is a parabola branches down. It has a maximum at a point `x_0=-(b)/(2a)=-1000/(2*(-2))=250` (ft). The corresponding `y_0=1000-2*250=500` (ft).
The answer: the bounding walls will have a length of 500 ft and the separation walls will have a length of 250 ft.
I believe you can now solve this problem for corrals with a different mutual arrangement.
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